جواب تمرین صفحه 132 درس 6 ریاضی دوازدهم تجربی (هندسه)

الف 

\(V - {V_1} = \frac{1}{3}\pi {(3)^2} \times 6 - \frac{1}{3}\pi {(1)^2} \times 2 = \frac{{52}}{3}\pi \)

ب ذوزنقه

\

\(S = \frac{1}{2}(2 + 6) \times 4 = 16\)

الف 

حجم استوانه کوچک – حجم استوانه بزرگ = حجم شکل

\(\begin{array}{l} = \pi \times {5^2} \times 3 - \pi \times {2^2} \times 3 = 75\pi - 12\pi \\ = 63\pi \end{array}\)

ب سطح مقطع موازی با قاعده یک دایره توخالی شعاع خارجی 5 و شعاع داخلی 2 است.

\(V = 2(\frac{\pi }{3} \times {2^2} \times 3) = 8\pi \)

الف

\(\begin{array}{l}F\left| \begin{array}{l}1\\3\end{array} \right.\\\\F'\left| \begin{array}{l}1\\ - 5\end{array} \right.\\\\O\left| \begin{array}{l}\frac{{1 + 1}}{2} = 1\\\frac{{3 + ( - 5)}}{2} = - 1\end{array} \right.\end{array}\)

\(x = 1\) معادله قطر بزرگ

\(y = - 1\)  معادله قطر کوچک

\(FF' = \left| {3 - ( - 5)} \right| = 8 = 2c \Rightarrow c = 4\)

ب

\(\begin{array}{l}{c^2} = {a^2} - {b^2} \Rightarrow {b^2} = {a^2} - {c^2} = 36 - 16 = 20\\\\ \Rightarrow b = \sqrt {20} = 2\sqrt 5 \Rightarrow BB' = 4\sqrt 5 \end{array}\)

خروج از مرکز:

\(e = \frac{c}{a} = \frac{4}{6} = \frac{2}{3}\)

الف

\(\begin{array}{l}e = \frac{c}{a} = \frac{4}{5} \Rightarrow c = \frac{4}{5}a\;\\\\BB' = 2b = 6 \Rightarrow b = 3\\\\{c^2} = {a^2} - {b^2} \Rightarrow \frac{{16}}{{25}}{a^2} = {a^2} - 9\\\\ \Rightarrow {a^2} = 25 \Rightarrow a = 5 \Rightarrow c = 4\end{array}\)

ب

\(\begin{array}{l}AA' = 2a = 10\\\\ \Rightarrow F\left| \begin{array}{l} - 4 + 4 = 0\\ - 1\end{array} \right.\;\;\;\;F'\left| \begin{array}{l} - 4 - 4 = - 8\\ - 1\end{array} \right.\\\\A\left| \begin{array}{l} - 4 + 5 = 1\\ - 1\end{array} \right.\;\;\;\;\;A'\left| \begin{array}{l} - 4 - 5 = - 9\\ - 1\end{array} \right.\\\\B\left| \begin{array}{l} - 4\\ - 1 + 3 = 2\end{array} \right.\;\;\;\;B'\left| \begin{array}{l} - 4\\ - 1 - 3 = - 4\end{array} \right.\end{array}\)